Fourier Transforms

Question 1

sinc(x) =

Answer 1

sinx/x

Question 2

proof of fourier transform

Answer 2

rk = r2π/T = ωr

Δω = 2π/T

1/T = Δω/2π

substitute into complex fourier series.

Let T -> inf

Σ Δω -> inf ∫ -inf

Question 3

Properties of the Fourier Transform

Answer 3

FT twice almost the original function

Linearity of FT

Scaling

Translation

Differentiation

Integration

Question 4

Scaling of Fourier Transform

Answer 4

a > 0 t/a -> t’ dt = adt t = at’

= a/√2π (inf ∫ -inf) dt’ f(at’) exp[-iωt]

f(ω/a) = 1/√2π (inf ∫ -inf) dt f(at) exp[-iωt]

Question 5

Translation of Fourier Transform

Answer 5

t+a = t’ dt = dt’

1/√2π (inf ∫ -inf) dt f(t’) exp[-iω(t’-a)]

= exp[iωa] 1/√2π (inf ∫ -inf) dt f(t) exp[-iωt]

apply fourier transform to both sides.

F[f(x-a)] = 1/√2π (inf ∫ -inf) dt f(t’) exp[-iω(t’+a)]

Question 6

Solving Differential Equations using Fourier Transforms

Answer 6

1) Differentiate Eq
2) FT of both sides
3) Exploit Linearity
4) Exploit Derivative
5) Rearrange for f(ω) and take the inverse

Question 7

FT of a gaussian is

Answer 7

another gaussian

Question 8

taking the fourier of sine and cosine

Answer 8

convert exponential to cos(ωt) - isin(ωt)

Question 9

Scaling property of the delta dirac function

Answer 9

(inf ∫ -inf) f(t) δ(at) dt = (inf ∫ -inf) f(t’/a) δ(t’) dt’/a

= 1/a f(0)

= 1/a (inf ∫ -inf) f(t) δ(t) dt

= (inf ∫ -inf) f(t) δ(t)/a dt

identify δ(at) = δ(t)/a

then do for opposite limit sign

giving δ(at) = - δ(t)/a

Question 10

for higher derivatives - delta dirac

Answer 10

(inf ∫ -inf) dt f(t) δ^(n) t = (-1)^(n) f^(n) (0)

Question 11

Fourier transform of the delta function

Answer 11

f(t) = 1/2π (inf ∫ -inf) du (inf ∫ -inf) dv f(v) exp[iω(t-u)]

(inf ∫ -inf) dv f(v) δ(u-t) = f(t)

δ(t) = 1/2π (inf ∫ -inf) dω exp[iωt]

δ(t) = 1/√2π (inf ∫ -inf) 1/√2π exp[iωt] dω

δ(tilda)(ω) = 1/√2π

Question 12

For the fourier transform x -> ∞

Answer 12

The FT exists if the integral is finite for all times t

therefore f(x) -> 0 as x -> ∞

Question 13

Fourier domain PDE

Answer 13

take fourier transform of both sides

the PDE is now an ODE

take FT of initial conditions for general solution

Question 14

convolution applications

Answer 14

in photography

the measurements are given by the convolution of the signal and resolution function